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Question

Find the sum of the first n terms of the series :3+7+13+21+31+...

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Solution

The given series is S=3+7+13+21+31+....

S=3+7+13+21+31+...an1+an

S=3+7+13+21+31+...an2+an1+an

On subtracting both the equation, we obtain

SS=[3+(7+13+21+31+...+an1+an)][(3+7+13+21+31+...+an1+an]

SS=3+[(73)+(137)+(2113)+...(an+an1)]an

0=3+[4+6+8+...(n1)terms]an

an=3+[4+6+8+...(n1)terms]

an=3+(n12)[2×4+(n11)2]

=3+(n12)[8+(n2)2]

=3+(n12)(2n+4)

=3+(n1)(n+2)

=3+(n2+n2)

=n2+n+1

S=nk=1ak=nk=1k2+nk=1k+nk=11

=n(n+1)(2n+1)6+n(n+1)2+n [by using ni=1k2=12+22+32++n2=n(n+1)(2n+1)6]

=n[(n+1)(2n+1)+3(n+1)66]

=n[2n2+3n+1+3n+3+66]

=n[2n2+6n+106]

=n3(n2+3n+5)

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